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# I really need help ASAP. Thanks.

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Points $A$, $B$, and $C$ are on sides $\overline{WX}$, and $\overline{YZ}$, and $\overline{XY}$ of rectangle $WXYZ$ as shown below such that $XA=4$, $YB = 18$, $\angle ACB= 90^\circ$, and $YC = 2XC$. Find $AB$. That is the best latex I could come up with. :p. Sorry.

https://latex.artofproblemsolving.com/0/6/1/061230f0b1b17bcd40886b56dbcbbc0368db13b5.png

May 3, 2020
edited by Guest  May 3, 2020
edited by Guest  May 3, 2020
edited by Guest  May 3, 2020

#4
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Points $$A$$, $$B$$, and $$C$$ are on sides $$\overline{WX}$$, $$\overline{YZ}$$, and $$\overline{XY}$$ of rectangle $$WXYZ$$ as shown below such that
$$XA = 4$$, $$YB = 18$$$$\angle ACB= 90^\circ$$, and $$YC = 2XC$$.
Find $$AB$$. $$\begin{array}{|rcll|} \hline \tan(A) = \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{XC}{9} \\\\ 36 &=& XC^2 \\\\ 6 &=& XC \\ \mathbf{XC} &=& \mathbf{6} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline x^2 &=& (3XC)^2 + BD^2 \quad | \quad BD =YB-XA \\ x^2 &=& (3XC)^2 + (YB-XA)^2 \\ x^2 &=& (3*6)^2 + (18-4)^2 \\ x^2 &=& 18^2 + 14^2 \\ x^2 &=& 324 + 196 \\ x^2 &=& 520 \\ x^2 &=& 4*130 \\ x &=& 2\sqrt{130} \\ \mathbf{x} &=& \mathbf{22.8035085020} \\ \hline \end{array}$$

$$\mathbf{AB \approx 22.8}$$ May 4, 2020
edited by heureka  May 4, 2020

#1
+1   May 3, 2020
#4
+2

Points $$A$$, $$B$$, and $$C$$ are on sides $$\overline{WX}$$, $$\overline{YZ}$$, and $$\overline{XY}$$ of rectangle $$WXYZ$$ as shown below such that
$$XA = 4$$, $$YB = 18$$$$\angle ACB= 90^\circ$$, and $$YC = 2XC$$.
Find $$AB$$. $$\begin{array}{|rcll|} \hline \tan(A) = \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{XC}{9} \\\\ 36 &=& XC^2 \\\\ 6 &=& XC \\ \mathbf{XC} &=& \mathbf{6} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline x^2 &=& (3XC)^2 + BD^2 \quad | \quad BD =YB-XA \\ x^2 &=& (3XC)^2 + (YB-XA)^2 \\ x^2 &=& (3*6)^2 + (18-4)^2 \\ x^2 &=& 18^2 + 14^2 \\ x^2 &=& 324 + 196 \\ x^2 &=& 520 \\ x^2 &=& 4*130 \\ x &=& 2\sqrt{130} \\ \mathbf{x} &=& \mathbf{22.8035085020} \\ \hline \end{array}$$

$$\mathbf{AB \approx 22.8}$$ heureka May 4, 2020
edited by heureka  May 4, 2020